The time period of a simple pendulum is T. When the length is increased by 10 cm, the period is $T_{1}$ .When the length is decreased by 10 cm, its periodis T $_{2}$ . Then relation between T, T $_{1}$ , T $_{2}$ is

2 / T_{2} = 1 / T_{1}^{2} + 1 / T_{2}^{2}

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2 / T_{2} = 1 / T_{1}^{2} - 1 / T_{2}^{2}

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2 T^{2} = T_{1}^{2} + T_{2}^{2}

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2 T^{2} = T_{1}^{2} - T_{2}^{2}

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Solution

The correct option is B $2 T^{2} = T_{1}^{2} + T_{2}^{2}$
$T = 2 π \sqrt{\frac{l}{g}}$
$T_{1} = 2 r \sqrt{\frac{l + 10 c m}{g}}; T_{2} = 2 π \sqrt{\frac{l - 10 c m}{g}}$
$T {_{1}}^{2} + T {_{2}}^{2} = 4 π^{2} (\frac{2 l + 10 c m - 10 c m}{g})$
$T {_{1}}^{2} + T {_{2}}^{2} = 2 T^{2}$

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The time period of a simple pendulum is T. When the length is increased by 10 cm, the period is T1.When the length is decreased by 10 cm, its periodis T2. Then relation between T, T1, T2 is

The correct option is B 2T2=T21+T22T=2π√lgT1=2r√l+10cmg;T2=2π√l−10cmgT12+T22=4π2(2l+10cm−10cmg)T12+T22=2T2

The time period of a simple pendulum is T. When the length is increased by 10 cm, the period is $T_{1}$ .When the length is decreased by 10 cm, its periodis T $_{2}$ . Then relation between T, T $_{1}$ , T $_{2}$ is

The correct option is B $2 T^{2} = T_{1}^{2} + T_{2}^{2}$
$T = 2 π \sqrt{\frac{l}{g}}$
$T_{1} = 2 r \sqrt{\frac{l + 10 c m}{g}}; T_{2} = 2 π \sqrt{\frac{l - 10 c m}{g}}$
$T {_{1}}^{2} + T {_{2}}^{2} = 4 π^{2} (\frac{2 l + 10 c m - 10 c m}{g})$
$T {_{1}}^{2} + T {_{2}}^{2} = 2 T^{2}$